Workers were given training for two different methods ofbuilding a widget with t

Question

Workers were given training for two different methods ofbuilding a widget with the results listed in the table below.Table

Worker=                  1 through 7

Time with Method 1 = 6.2, 7.0, 9.5, 5.8, 6.7, 8.3, 7.5

Time with Method 2 = 6.8, 7., 8.7, 7.1, 7.3, 8.1, 7.8

Difference     =      -.6, -.1, .8, -1.3, -.6, .2 -.3


Please answer the following questions. Show all equations and allwork. Round all answers to 3 decimal places.
a. Calculate the sample mean difference ??
b. Calculate the sample standard deviation ????
c. How many degrees of freedom are there?
d. Calculate the test statistic for a hypothesis test for adifference between the means.
e. Is there enough evidence to show there is a difference betweenthe two methods with =.05? Please support your answerstatistically.

Explanation / Answer

mean difference is -.2714 sample std. deviation is .6676 degrees of freedom is n-1=7-1=6 test statistic=t=-1.6/(sample standarddeviation/n)=-1.6/(.6676/7)=-1.07566 The p-value is .3234 (found using software), so at =.05,we do not reject null, and cannot conclude that there is adifference between the two methods.

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